Sin(3x)=3sinx ?

This is a bizarre little proof that is probably wrong, but I want to know where
[tex]\lim_{\theta\rightarrow0}\frac{\sin{3\theta}}{\theta} = 3[/tex]
This is provable quite easily, I don't think I need to do that atm...
well, we know that
[tex]\lim_{\theta \rightarrow 0} \frac{\sin{\theta}}{\theta} = 1[/tex]
So that would imply that
[tex]3\lim_{\theta \rightarrow 0} \frac{\sin{\theta}}{\theta} = 3[/tex]
thus implying that
[tex]\lim_{\theta \rightarrow 0} \frac{3\sin{\theta}}{\theta} = 3[/tex]
By substitution
[tex]\lim_{\theta \rightarrow 0} \frac{\sin{3\theta}}{\theta} = \lim_{\theta \rightarrow 0} \frac{3\sin{\theta}}{\theta}[/tex]
(iffy step)
If you drop the limit on both sides...
[tex]\frac{\sin{3\theta}}{\theta}=\frac{3\sin{\theta}}{\theta}[/tex]
Which oh so quickly becomes
[tex]\sin{3\theta}=3\sin{\theta}[/tex]

Yes, that step is "iffy". With it, I can "prove" that sin(3θ) equals any function with the same limit as θ approaches zero. The step is invalid for the same reason that it is invalid to conclude that sin(3θ)=sin(θ) just because those two functions happen to be equal at θ=0.